At any instant the velocity of a particle of mass $500\,g$ is $\left(2 t \hat{ i }+3 t ^2 \hat{ j }\right)\,ms ^{-1}$. If the force acting on the particle at $t=1 s$ is $(\hat{i}+x \hat{j}) N$. Then the value of $x$ will be:

A particle of mass $ 0.3 \,kg$ is subjected to a force $F = – kx$ with $k = 15\,N/m.$ What will be its initial acceleration if it is released from a point $20 \,cm$ away from the origin ………. $m/s^2$

A force $\vec{F}=(40 \hat{i}+10 \hat{j})\, N$ acts on a body of mass $5\, {kg}$. If the body starts from rest, its position vector $\vec{r}$ at time $t=10\, {s}$, will be –

A force $F = Be^{-Ct}$ acts on a particle whose mass is $m$ and whose velocity is $0$ at $t = 0$. It’s terminal velocity is :

A particle of mass $M$ originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation

$F=F_{0}\left(1-\left(\frac{t-T}{T}\right)^{2}\right)$

Where $F_{0}$ and $T$ are constants. The force acts only for the time internal $2 T$. The velocity $v$ of the particle after time $2 {T}$ is –